Dynamics Fem

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Question 1  The weak form fo r transient linear elasticity in one dimension is as follows: Introduce w, an arbitrary weight function and u, a trial solution such that: w = { w w  H 1 , w (  Γ  D ) =0 } u= { uu  H 1 ,u (  Γ  D ) u } Multiplying by w and integrating over the domain wρA ´ u (  x ,t ) dx = w ( p (  x , t ) x  +b (  x , t ) ) dx wρA  ∂ 2 u t 2  dx = w (  EA  ∂ 2 u x 2  +b ) dx wρA  ∂ 2 u t 2  dx w (  EA  ∂ 2 u x 2 ) dx = wbdx Applying integration by parts on the term w (  EA  ∂ 2 u x 2 ) dx wρA  ∂ 2 u t 2  dx wEA  ∂ u dx  ¿  Γ  N + w x  (  EA  ∂ u x ) dx = wbdx trong !orm:  ρA ´ u (  x , t )= p (  x ,t ) x  + b (  x ,t )  p (  x ,t )=  EA  ∂ u (  x , t ) x BC : u uon Γ  D ; p  p on Γ  N 

Transcript of Dynamics Fem

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"ynamics # $ro%ect 1 August &'1(

∫wρA ∂

2u

∂ t 2 dx +∫

∂w

∂ x ( EA ∂u∂x )dx=∫wbdx+w ´ p

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"ynamics # $ro%ect 1 August &'1(

Question &

A )arlerkin appro*imation was used on the weak form for transient linear elasticity

+shown in Question 1- This was used to discreti.e the problem in space, therefore

general e/uations for the relevant matrices for each of the elements is:

 Two di0erent ewmark time schemes were implemented to solve the initial

boundary problem over a period of time- The general e*pressions for these time

schemes are:

 The results using consistent mass matrices are as follows:

a Trape.oidal 2ule +  β=

1

4, γ =

1

2¿

 M e= ρ

e A

el

e∫ N 

T  Nd Ω

e

 K e= E

e A

e

le   ∫B

T BdΩ

e

 F e=   N 

T  Fd Ω

e

tart from:

 M un+1+C un+1+ K un+1= F n+1

Assume e/uilibrium at n31

vn=un , n=un

un+1=un+! t vn+! t 2

2[ (1−2 β ) n+2 β n+1]

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"ynamics # $ro%ect 1 August &'1(

 The trape.oidal rule seems to be unconditionally stable- Though a coarse

discreti.ation of time results in a poor value, the numbers are not growing

e*ponentially nor running away as the re4nement decreases- The di0erence

in the deflection of the bar when decreasing 5t by 1'' is appro*imately '-6 *

1'7(, which is &'8 error in using a coarse discreti.ation-

b 9entral di0erence method +  β=0 , γ =

1

2¿

In comparison to the trape.oidal rule, the central di0erence method is

conditionally stable- This means that there is a ma*imum value for 5t, where

any larger 5t value will cause instability in the numerical implementation- At

5t 1-' * 1'7; s, the graph was still not generating and A values were

found in the matri* for the displacement- It is only around 5t < ;-= * 1'7>

s,that a valid solution appear-

 The mass matrices were then lumped using the row7sum techni/ue-

c Trape.oidal 2ule +  β=

1

4, γ =

1

2¿

!igure &: 5t '-''''1s!igure 1: 5t '-''1s +?steps

!igure (: 5t ;-== * 1' 7> s +?steps!igure @: 5t '-''1s +?steps

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"ynamics # $ro%ect 1 August &'1(

 The lumped mass matrices and the consistent mass matrices results were

very similar- This shows that there is no di0erence between using a di0erent

mass matri* whilst using the trape.oidal rule-

d 9entral di0erence method +  β=0 , γ =1

2¿

 The critical time step for the central di0erence method had increased when

the mass matrices were lumped- It is more ecient to use a lumped mass

matri* when using 9"M since fewer time steps are re/uired to converge to a

good solution-

Bverall comments about the shape of the graph of a good solution shows that the

last element +from C '-D; # 1m, the deEection is constant- This is because thelast element has a much greater sti0ness than the rest of the bar- Also, only half of 

the bar e*periences deEection +the half closest to the end sub%ected to the

concentrated and compressive end load- This can be due to the fact that the bar is

a 1m long and the force is relatively small-

!igure ;: 5t '-''1s +?steps !igure >: 5t '-''1s +?steps

!igure 6: 5t 1-11 * 1' 7; s +?steps !igure =: 5t 1-' * 1' 7; s +?steps

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"ynamics # $ro%ect 1 August &'1(

Question @

 The bottom7left /uarter of a 1m * 1m s/uare plate that underwent transient thermal

conduction was modelled- The governing e/uation for transient thermal conduction

and the relevant elemental matrices for this particular are:

 The problem is a parabolic e/uation- Fackward Guler was the chosen time scheme

used to model the problem because it is unconditionally stable however the MATCAF4le allows for this to be changed- The general e*pression for this time scheme is:

 The results are as follows:

 M  T + KT = F 

 K e=" ∫B

T Bd Ω

e

 M e= ρ

e# p∫ N 

T  Nd Ω

e

 F e=∫ N 

T $d Ω

e

 M un+1+ K un+1= F n+1

un+1=un+! t v n+% 

!igure 1':!igure D:

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"ynamics # $ro%ect 1 August &'1(

 The heat evolution of the plate that has an initial temperature of 'H and a heat

source of s f oe7t  increases to a ma*imum of appro*imately '-(@ H at 37'-=s-

 Thereafter, the temperature shown in the graph seems to decay e*ponentially- Thisis to be e*pected as the heat source is e*ponentially decaying as time goes by-

!igure1&:!igure11: